let L be non empty satisfying_DN_1 ComplLLattStr ; :: thesis: for x, y, z being Element of L holds (x `) + (((y + x) + z) `) = x `
let x, y, z be Element of L; :: thesis: (x `) + (((y + x) + z) `) = x `
set X = x ` ;
(x `) + (((y + ((x `) `)) + z) `) = x ` by Th42;
hence (x `) + (((y + x) + z) `) = x ` by Th23; :: thesis: verum